Existence and Stability of Implicit Fractional Differential Equations with Stieltjes Boundary Conditions Involving Hadamard Derivatives

In this article, we make analysis of the implicit fractional differential equations involving integral boundary conditions associated with Stieltjes integral and its corresponding coupled system. We use some sufficient conditions to achieve the existence and uniqueness results for the given problems by applying the Banach contraction principle, Schaefer’s fixed point theorem, and Leray–Schauder result of the cone type. Moreover, we present different kinds of stability such as Hyers–Ulam stability, generalized Hyers–Ulam stability, Hyers–Ulam–Rassias stability, and generalized Hyers–Ulam–Rassias stability by using the classical technique of functional analysis. At the end, the results are verified with the help of examples.


Introduction
Fractional-order derivatives are the generalization of integer-order derivatives.e idea of fractional-order derivatives has been introduced at the end of the sixteen century when Leibniz used the notation (d α /dσ α ) for α th -order derivative.It is well known that if ϕ(σ) � (1/2)σ 2 , then ϕ ′ (σ) � σ and ϕ ″ (σ) � 1, respectively.But, if the order is in the fractional form, i.e., α � (1/2), α � (1/3), and so on, then what would be the result? is question was mentioned in a letter to L'Hospital by Leibniz in (1695) [1].Since then, several mathematicians such as Fourier, Laplace, and Letnikov contributed to the development of the fractional calculus.Riemann and Liouville had worked on this problem and introduced the Riemann-Liouville fractional derivative, which was further generalized by Caputo and can be considered as the fundamental concept in fractional calculus.Fractional calculus has many applications in many scientific disciplines, e.g., in the fields of signal and image processing [2], mathematical biological systems [3], electronics, economics, control theory [4], chemistry, biophysics, and blood flow phenomena.For more applications of fractional differential equations, we refer the reader to [5,6] and the references cited therein.
In 1892 [7], Hadamard presented a concept of fractional derivative, which is different from Caputo-and Riemann-Liouville-type fractional derivatives.An important feature of the Hadamard fractional derivative is that it contains a logarithmic function of an arbitrary exponent in its definition.Here, we stress that the studies about Hadamard fractional differential equations are still at the early stage and need additional analysis.For more details and recent contributions to the topic, see [8][9][10][11].
In the field of fractional differential equations, the area which received great attention from researchers is the existence of solutions of different boundary-value problems.Many researchers used different fixed-point theorems and developed different approaches for the existence of solutions of complicated boundary-value problems.For details, we refer the reader to [12][13][14].e study of coupled systems of differential equations is also very significant because these types of systems appear naturally in many problems of applied nature.For details and examples, the reader may see [15][16][17][18] and the references cited therein.
In this article, we extend the system of Ren and Zhai [30] to implicit Hadamard fractional derivatives having the Stieltjes integral condition instead of the Riemann-Liouville fractional q-derivative and present its existence, uniqueness, at least one solution, and different kinds of Ulam's stabilities.We study the following system: where α ∈ (2, 3], ϕ: involving the Stieltjes integral with respect to the function μ: [1,2] ⟶ R, μ(σ) is right continuous on [1,2), left continuous at σ � 2. Particularly, μ is a nondecreasing function with μ(1) � 0; then, dμ is a positive Stieltjes measure.We also extend system (5) to the coupled system and discuss its existence, uniqueness, at least one solution, and different kinds of Ulam's stabilities of the problem: where involving the Stieltjes integral with respect to the function μ: Particularly, μ is a nondecreasing function with μ(1) � 0; then, dμ is a positive Stieltjes measure.e rest of the article is arranged as follows: In Section 2, we present some basic definitions, lemmas, and theorems that are used in our main results.In Section 3, we use different conditions and some standard fixed-point theorems for the existence and uniqueness of solutions to the given system (5) and its corresponding coupled system (7).In Section 4, we present Ulam's stabilities for the given systems ( 5) and (7) under some specific conditions.At the end, examples are given to illustrate the main results.
roughout the paper, we assume that (M 6 ): let Ψ ∈ C(J, R) be an increasing function; then, there is Ω Ψ > 0 such that, for each σ ∈ J, the inequality holds.

Preliminaries
In this section, we present some useful definitions, lemmas, and theorems, which will be used throughout the manuscript\enleadertwodots.
Theorem 1 (see [32]) (Arzela-Ascoli's eorem).Let B ⊂ C(J, R) be relatively compact, and (A) B is a uniformly bounded set such that there exists ϖ > 0 with (B) B is an equicontinuous set, i.e., for every ϵ > 0, there exists δ > 0 such that, for any Theorem 2 (see [33]) (Banach Fixed-Point eorem).Let B be a nonempty closed subset of a Banach space X. en, any contraction mapping T from X into itself has a unique fixed point.
Theorem 3 (see [33]) (Schaefer's Fixed-Point eorem).Let X be a Banach space.Suppose the operator T: X ⟶ X is a continuous compact mapping (or completely continuous).Moreover, suppose is a bounded set.en, B has at least one fixed point in X.
Similarly, the norm defined on the product space Theorem 4 (see [34]).Let X be a Banach space containing a cone

Existence and Uniqueness
In this section, we give the existence and uniqueness of solutions of (5) and its coupled system (7).

Existence and Uniqueness Solution for System (5).
Our first result is stated as follows.
(3) Since we know that Green's function of the considered problem is in the form using (M 5 ) and mean-value theorem [35] with ρ ∈ [1, 2] and μ(2) � P > 0, we get Hence, the proof of (3) is complete.
If u is the solution of the given system (5) and σ ∈ J, then Complexity where We define an operator T: X ⟶ X as □ Theorem 5. Let (M 2 ) − (M 5 ) hold.en, the operator T, defined in (45), is compact.
Proof.To show that the operator T is compact, we follow several steps.
Step 1: we consider a sequence u n   such that u n ⟶ u in X; then, for each σ ∈ J, we have 10 Complexity From (46), we can write Now, by (M 4 ), we have which implies Since we supposed that u n ⟶ u, z n ⟶ z as n ⟶ ∞ for each σ ∈ J. So, by Lebesgue dominated convergence theorem [36], (47) gives Hence, T is continuous.
Step 2: now, we are going to prove that the operator T is bounded in set X.For this, we show that, for any � p > 0, there exists � q > 0 such that, for each we have From ( 45), for each σ ∈ J, we have Now, by (M 3 ) and (46), we have Taking sup σ>0 , we get (58)

Complexity
Hence, T(Q) is uniformly bounded.
Step 3: now, to show that the operator T is equicontinuous in X.For this, let σ 1 , σ 2 ∈ J with σ 1 > σ 2 , since Q is a bounded set in X, and let u ∈ Q. en, e right-hand side of (59) approaches to zero as σ 1 ⟶ σ 2 .Hence, T(Q) is equicontinuous.As a consequence of Step 1 to 3, the operator T is completely continuous.erefore, in view of the Arzelà-Ascoli theorem, the operator T is compact.
□ Theorem 6.Let the hypotheses (M 3 ) and (M 5 ) hold, and if N < 1, the given problem ( 5) has at least one solution in X.
Proof.For the proof of this theorem, we are considering a set B ⊂ X which is defined in the following form: 12 Complexity We have to show that the set B is bounded.Let u ∈ X such that u(σ) � δTu(σ), where 0 < δ < 1. (61) en, for each σ ∈ J, we have Now, by (M 3 ), we have So, we get Putting (64) in (62) and taking sup σ>1 , we get Complexity 13 For simplicity, let So, (65) becomes is shows that the set B is bounded.So, by eorems 3 and 5, we get that the operator T has at least one fixed point.
erefore, the given problem ( 5) has at least one solution in X.

□
Theorem 7. Suppose that the hypothesis (M 2 ), (M 4 ), and (M 5 ) hold.en, the given problem ( 5) has a unique solution in X if Proof.We shall use the Banach contraction principle to prove that the operator T has a unique fixed point, which will be the unique solution of the given system (5), by considering the operator T: X ⟶ X defined in (45).Let u, u be the solution of (5), and for σ ∈ J, we have Now, by (M 4 ), we have which implies So, (70) becomes Using (M 5 ) and taking sup σ>1 on both sides, we get is implies that Hence, the operator T is a contraction.us, by the Banach contraction principle, we get that T has a unique fixed point, which is a unique solution of the given problem (5).□ (7).In this section, we show the existence and uniqueness of the solution of the system (7).First, we have the following: Complexity Lemma 5.

Existence and Uniqueness Solution for System
e system has a solution (u, v) if and only if where Proof.
e proof is similar to that given in Lemma 3 and, hence, is not included here.We use the following notations for convenience: Hence, for σ ∈ J, (78) becomes where y, z ∈ X satisfying the functional equation.
)) of the system (7) have the following properties: ( where Proof. (1) It is easy to prove that G α,c (σ, η) is continuous, so we leave it.
(2) Since we know that Green's functions of the considered problem (7) is in the form using (M 5 ) and the mean value theorem [35] with ρ ∈ [1, 2] and μ(2) � P > 0, we get Similarly, we can obtain 18 Complexity Hence, the proof of 2 is complete.
If u, v are the solutions of the given system (7) and σ ∈ J; then, Now, we transform the given system (7) into a fixedpoint problem.Let an operator T: X × X ⟶ X × X be defined as Complexity en, the solution of ( 7) coincides with the fixed point of T, where en, the operator T: 87) is completely continuous.
Proof.In view of continuity of ϕ 1 , ϕ 2 and G α,c (σ, η), T is also continuous for all (u, z) ∈ C � .Suppose � B⊆C � is a bounded set.So, for every u ∈ � B, we have (90) So, we obtain Now, using 2 of Lemma 6, (M 11 ), and (91) in (89), we get In the same way, we obtain us, from (92) and (93), we get Complexity us, T is uniformly bounded.Now, we prove the operator T is equicontinuous.For this, suppose σ 1 > σ 2 ∈ J and u, v ∈ � B; then, 22 Complexity In the same way, we can show that e right-hand sides of ( 94) and ( 96) approache to zero as σ 1 ⟶ σ 2 .Hence, by the Arzelà-Ascoli theorem, T is equicontinuous and uniformly equicontinuous.Also, it is very easy to prove that T( � B) ⊂ � B. erefore, T is completely continuous.
, and we consider where (99) Now, using (M 10 ), Complexity 23 Substituting (100) in (98) and taking sup σ∈J , we get (101) In the same way, we can obtain So, from ( 101) and ( 102), we get us, T is a contraction.erefore, by the Banach contraction principle, T has a fixed point.So, we infer that the given coupled system (7) has a unique solution.□ Theorem 10.In view of the continuity of the functions ϕ 1 , ϕ 2 and supposing (M 9 ) and (M 13 ) with hold, the coupled system ( 7) has at least one solution.
Proof.Let a set � B be defined as Furthermore, the operator defined by T: Similarly, erefore, 24 Complexity so T(u, z) ∈ � B .us, in view of eorem 8, T: � B ⟶ � B is completely continuous.Now, we consider an eigenvalue problem defined as So, in view of the solution (u, v) of (109), we obtain Similarly, us, From equation (112), we get (u, v) ∉ z � B. So, in view of eorem 4, T has at least one fixed point which lies in � B . is shows there is at least one solution of the coupled system (7).

Hyers-Ulam Stability
In this section, we provide novel characterizations of the Hyers-Ulam stability for systems (5) and (7).For the various concepts of Hyers-Ulam stability, see, for example, [37].

Hyers-Ulam Stability Concepts for System (5) Definition 3.
e problem ( 5) is said to be Hyers-Ulam stable if there exists some constant K > 0 such that, for any ϵ > 0 and for any solution u ∈ X of the inequality there exists a solution  u ∈ X of (5) with Definition 4. e problem ( 5) is said to be generalized Hyers-Ulam stable if there exists Ψ ∈ C(R + , R + ) with Ψ(0) � 0 such that, for any solution u ∈ X of the inequality (113), there exists a solution  u ∈ X of (5) satisfying Definition 5. e problem ( 5) is said to be Hyers-Ulam-Rassias stable with respect to Ψ ∈ C(J, R) if there exists some constant K Ψ > 0 such that, for any ϵ > 0 and for any solution u ∈ X of the inequality there exists a solution  u ∈ X of (5) with Definition 6. e problem ( 5) is said to be generalized Hyers-Ulam-Rassias stable with respect to Ψ ∈ C(J, R) if there exists some constant K Ψ > 0 such that, for any solution u ∈ X of the inequality (116), there exists a solution  u ∈ X of (5) satisfying Remark 2. Let u ∈ X be a solution of the inequality (113); then, there exists a function ) is the solution of the inequality (113), then u will be the solution of the following integral inequality: Proof.Let u be the solution of the inequality (113).So, in view 2 of Remark 2, we have So, for σ ∈ J, the solution of (120) will be in the form From equation (121), we have For computational convenience, we use ω 1 (σ) for the sum of terms which are free of U ϕ , so we have From above, we have Using (3) of Lemma 4 and 1 of Remark 2, we get □ Theorem 11.Under the hypothesis (M 2 ), (M 4 ), and (M 5 ) and if holds, then the given system ( 5) is HUS and, consequently, GHUS.
Proof.Let u ∈ C(J, R) be the solution of (113) and  u be the unique solution of the system given by en, for σ ∈ J, the solution of (127) is Using Lemma 7 in (129), we have Complexity where z u , z  u ∈ C(J, R) are of the form By (M 4 ), we get which implies Using (3) of Lemma 4 and ( 133) in (130), we get where us, problem (5) is HUS.
Lemma 8. Let the hypothesis (M 6 ) hold, and suppose u ∈ C(J, R) is the solution of the inequality (116); then, u is a solution of the following inequality: Proof.From Lemma 7, we have By using (3) of Lemma 4, 1 of Remark 2, and (M 6 ), we get □ 28 Complexity Theorem 12.Under the hypothesis (M 2 ) and (M 4 )-(M 6 ) and if the inequality holds, then the given system ( 5) is stable in the sense of HUR.
Proof.Let u ∈ C(J, R) be the solution of (113) and  u be the unique solution of the system given by en, for σ ∈ J, the solution of ( 141) is Using (M 4 ) in a similar way as used in eorem 11, we get Now, by Lemma 8 and by (144), (143) becomes us, we have where Hence, the given system ( 5) is HUR stable.

Hyers-Ulam Stability Concepts for System
e given system (7) has HUS if there exists C α,c � max(C α , C c ) > 0 such that there exist some ε α,c � max(ε α , ε c ) > 0, and for every solution (u, v) ∈ X × X of the inequality there exists a solution Definition 8 (see [38]).e given system (7) has GHUS if there exists C ′ ∈ C(R + , R + ) with C ′ (0) � 0 such that, for any solution (u, v) ∈ X × X of the inequality (149), there exists a solution ( u,  v) ∈ X × X of (7) satisfying Definition 9 (see [38]).e given system (7) has HURS with respect to there exists a solution ( u,  v) ∈ X × X with Definition 10 (see [38]).e given system (7) has GHURS with respect to Remark 5. Let (u, v) ∈ X × X be a solution of the inequality (149), if there exist functions U ϕ 1 , U ϕ 2 ∈ C(J, R) depending on u, v, respectively, such that Let (u, v) ∈ X × X be the solution of (149); then, for σ ∈ J, we have Complexity Proof.By 2 of Remark 5 and for σ ∈ J, we have So, for σ ∈ J, the solution of (156) will be in the form From the first equation of system (157), we have (158) For computational convenience, we use ω 1 (σ) for the sum of terms which are free of U ϕ 1 , so we have So, from the above and taking the absolute value, (158) becomes Using 2 of Lemma 6 and 1 of Remark 5, we get In the same way, we have □ Theorem 13.Under the hypothesis (M 10 ) and if holds, then the given system ( 7) is stable in the sense of HU.
Proof.Let (u, v) ∈ X × X be the solution of ( 149) and ( u,  v) ∈ X × X be the solution to the system Complexity en, for σ ∈ J, ( u,  v) is the solution of (164), i.e., where y, y  u ∈ X are of the form By (M 10 ), we get and we obtain Using 2 of Lemma 6 and (168) in (166), we get Similarly, we have where z, z  v ∈ X in the form We write (170) and (171) as In the matrix form, the abovementioned inequalities can be written as Solving the abovementioned inequality, we have Further simplification gives from which we have Let ε α,c � max ε α , ε c  ; then, from (178), we have where (180) □ Remark 6.By setting C ′ (ϵ α,c ) � C α,c ϵ α,c , C ′ (0) � 0 in (179), by Definition 8, the given system (7) is GHUS.
Remark 7.Under the hypothesis (M 12 ) and (163) and by using Definitions 9 and 10, one can repeat the process of Lemma 9 and the eorem 13, and the system (7) will be HUR and GHURS.

Complexity
Furthermore, K > 0 with condition (126) holds, and Hence, with the help of eorem 11, the given system (181) is HUS and, hence, GHUS.Also, by checking the conditions of eorem 12, it can be easily verified that the considered problem (181) is HURS and GHURS.(193) Hence, (188) has a unique solution.Furthermore, K > 0 with condition (126) holds, and Hence, with the help of eorem 11, the given system (188) is HUS and, hence, GHUS.Also, by checking the conditions of eorem 11, we can find that the considered problem (188) is HURS and GHURS.( By simple computations, we found that From eorem 9, we use the inequality which are found as Hence, (195) has a unique solution, and Hence. with the help of eorem 13, the given system (195) is HUS.Also, by using (M 12 ) and Remark 6 and 7, the given system is GHUS, HURS, and GHURS.

Conclusions
We have obtained some appropriate conditions for the existence, uniqueness, and Ulam's stabilities of the system (5) and its corresponding coupled system (7).e required results are obtained using the Banach contraction principle, Schaefer's fixed point theorem, Arzela-Ascoli theorem, and Leray-Schauder of the cone type.Also, with the help of some sufficient conditions, we have derived different kinds of Ulam's stabilities for the solution of the problem (5) and the coupled problem (7).Additionally, examples are given to support the main results.
are the continuous functions and H D α , H D β , H D c , and H D δ are the Hadamard fractional derivatives of orders α, c, respectively with